# evaluation of a double integral over the first quadrant

Evaluate $$\int_0^\infty \int_0^\infty \frac{\exp(-x-y)}{(x+y)}dxdy$$

I have shown that the integral is finite and bounded by $\frac \pi 2$.

## 2 Answers

Transforming to polar coordinates, we find that

\begin{align} I&=\int_0^\infty \int_0^\infty \frac{e^{-(x+y)}}{x+y}\,dx\,dy\\\\ &=\int_0^{\pi/2}\int_0^\infty \frac{e^{-r(\cos(\theta)+\sin(\theta))}}{\cos(\theta)+\sin(\theta)}\,dr\,d\theta\\\\ &=\int_0^{\pi/2}\frac{1}{(\cos(\theta)+\sin(\theta))^2}\,d\theta\\\\ &=\int_0^{\pi/2}\frac{1}{1+\sin(2\theta)}\,d\theta\\\\ &=\frac12 \int_0^\pi \frac{1}{1+\sin(\theta)}\,d\theta\\\\ &=1 \end{align}

Introduce the new variables $$u =x+y, \quad v= y.$$ The Jacobian of the variable transform is unity. The integral reads $$I = \int_0^\infty\!du\,\int_{0}^u\!dv\, \frac{e^{- u}}{ u}= \int_0^\infty\!du\,e^{-u} = 1.$$ The region of integration can be obtain from the set of inequalities $$0\leq x = u-v, \quad 0 \leq y = v$$ that is equivalent to $$0\leq u, \quad 0\leq v \leq u.$$